Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <cfloat>
#include <stack>
#include <queue>
#include <vector>
typedef long long int ll;
#define BIG_NUM 10000000000
#define MOD 1000000007
#define EPS 0.000001
using namespace std;

struct Info{
	Info(int arg_node_id,int arg_cost_sum){
		node_id = arg_node_id;
		cost_sum = arg_cost_sum;
	}
	int node_id,cost_sum;
};

int main(){

	int N,M,start,goal;
	scanf("%d %d %d %d",&N,&M,&start,&goal);

	ll* from_start_Dist = new ll[N+1];
	ll* from_goal_Dist = new ll[N+1];
	ll *Dist_num_from_start = new ll[N+1];
	ll *Dist_num_from_goal = new ll[N+1];

	vector<int> V[N+1];

	for(int i = 1; i <= N; i++){
		from_start_Dist[i] = BIG_NUM;
		from_goal_Dist[i] = BIG_NUM;
		Dist_num_from_start[i] = 0;
		Dist_num_from_goal[i] = 0;
	}

	from_start_Dist[start] = 0;
	from_goal_Dist[goal] = 0;
	Dist_num_from_start[0] = 0;
	Dist_num_from_goal[0] = 0;

	int from,to;

	for(int i = 0; i < M; i++){
		scanf("%d %d",&from,&to);
		V[from].push_back(to);
		V[to].push_back(from);
	}

	queue<Info> Q;

	for(int i = 0; i < V[start].size(); i++){
		from_start_Dist[V[start][i]] = 1;
		Q.push(Info(V[start][i],1));
	}

	int min_index;

	while(!Q.empty()){
		if(Q.front().cost_sum > from_start_Dist[Q.front().node_id]){
			Q.pop();
			continue;
		}else{
			min_index = Q.front().node_id;
			Q.pop();
		}

		for(int i = 0; i < V[min_index].size();i++){
			if(from_start_Dist[V[min_index][i]] > from_start_Dist[min_index] + 1){
				from_start_Dist[V[min_index][i]] =from_start_Dist[min_index] + 1;
				Q.push(Info(V[min_index][i],from_start_Dist[V[min_index][i]]));
			}
		}
	}

	for(int i = 1; i <= N; i++){
		if(from_start_Dist[i] != BIG_NUM){
			Dist_num_from_start[from_start_Dist[i]]++;
		}
	}
	int length = from_start_Dist[goal];


	for(int i = 0; i < V[goal].size(); i++){
		from_goal_Dist[V[goal][i]] = 1;
		Q.push(Info(V[goal][i],1));
	}

	while(!Q.empty()){
		if(Q.front().cost_sum > from_goal_Dist[Q.front().node_id]){
			Q.pop();
			continue;
		}else{
			min_index = Q.front().node_id;
			Q.pop();
		}

		for(int i = 0; i < V[min_index].size();i++){
			if(from_goal_Dist[V[min_index][i]] > from_goal_Dist[min_index] + 1){
				from_goal_Dist[V[min_index][i]] =from_goal_Dist[min_index] + 1;
				Q.push(Info(V[min_index][i],from_goal_Dist[V[min_index][i]]));
			}
		}
	}

	for(int i = 1; i <= N; i++){
		if(from_goal_Dist[i] != BIG_NUM){
			Dist_num_from_goal[from_goal_Dist[i]]++;
		}
	}

	ll ans = 0;

	for(int i = 0; i <= length-2; i++){
		ans += Dist_num_from_start[i]*Dist_num_from_goal[(length-2)-i];
	}

	printf("%lld\n",ans);

	return 0;
}